Document related concepts
no text concepts found
Transcript
```CHAPTER 6
Some discrete distributions
6.1. Examples: Bernoulli, binomial, Poisson, geometric distributions
Bernoulli distribution
A random variable
be a
X
such that
Bernoulli random variable
P(X = 1) = p
with parameter
Var X = p − p2 = p(1 − p).
We denote such a random variable by
P(X = 0) = 1 − p is said
p. Note EX = p and EX 2 = p,
and
to
so
X ∼ Bern (p).
Binomial distribution
A random variable X has a
k) = nk pk (1 − p)n−k .
binomial distribution
We denote such a random variable by
The number of successes in
n
with parameters
n
and
p
cumbersome calculations one can derive
X = Y1 + · · · + Yn ,
EX = EY1 + · · · + EYn = np.
P(X =
X ∼ Binom (n, p).
Bernoulli trials is a binomial random variable.
is binomial, then
if
EX = np. An easier way
Yi are independent
where the
After some
is to realize that if
X
Bernoulli variables, so
We have not dened yet what it means for random variables to be independent, but here we
mean that the events such as
(Yi = 1)
are independent.
Proposition 6.1
X := Y1 + · · · + Yn , where {Yi }ni=1 are independent Bernoulli random variables
parameter p, then
EX = np, Var X = np(1 − p).
Suppose
with
Proof. First we use the denition of expectation to see that
n
n
X
X
n i
n i
n−i
EX =
i
p (1 − p)
=
i
p (1 − p)n−i .
i
i
i=0
i=1
Then
81
82
6. SOME DISCRETE DISTRIBUTIONS
EX =
n
X
i
i=1
= np
n!
pi (1 − p)n−i
i!(n − i)!
n
X
i=1
(n − 1)!
pi−1 (1 − p)(n−1)−(i−1)
(i − 1)!((n − 1) − (i − 1))!
n−1
X
(n − 1)!
pi (1 − p)(n−1)−i
i!((n
−
1)
−
i)!
i=0
n−1
X n − 1
= np
pi (1 − p)(n−1)−i = np,
i
i=0
= np
where we used the Binomial Theorem (Theorem 1.1).
To get the variance of
X,
we rst observe that
2
EX =
n
X
i=1
EYi2 +
X
EYi Yj .
i6=j
Now
EYi Yj = 1 · P(Yi Yj = 1) + 0 · P(Yi Yj = 0)
= P(Yi = 1, Yj = 1) = P(Yi = 1)P(Yj = 1) = p2
n
2
2
using independence of random variables {Yi }i=1 . Expanding (Y1 + · · · + Yn ) yields n terms,
2
2
of which n are of the form Yk . So we have n − n terms of the form Yi Yj with i 6= j . Hence
Var X = EX 2 − (EX)2 = np + (n2 − n)p2 − (np)2 = np(1 − p).
Later we will see that the variance of the sum of independent random variables is the sum
of the variances, so we could quickly get Var X = np(1 − p). Alternatively, one can compute
E(X 2 ) − EX = E(X(X − 1)) using binomial coecients and derive the variance of X from
that.
Poisson distribution
A random variable
X
has the
Poisson distribution
P(X = i) = e−λ
We denote such a random variable by
X ∼ Pois(λ).
∞
X
i=0
so the probabilities add up to one.
with parameter
λi
.
i!
Note that
λi /i! = eλ ,
λ
if
6.1. EXAMPLES: BERNOULLI, BINOMIAL, POISSON, GEOMETRIC DISTRIBUTIONS
83
Proposition 6.2
Suppose
X
is a Poisson random variable with parameter
λ,
then
EX = λ,
Var X = λ.
EX =
∞
X
ie
i
−λ λ
i=0
i!
−λ
=e
∞
X
λi−1
λ
= λ.
(i
−
1)!
i=1
Similarly one can show that
2
E(X ) − EX = EX(X − 1) =
∞
X
i(i − 1)e−λ
i=0
= λ2 e−λ
∞
X
i=2
λi
i!
i−2
λ
(i − 2)!
2
=λ ,
so
EX 2 = E(X 2 − X) + EX = λ2 + λ,
Example 6.1.
and hence
Var X = λ.
Suppose on average there are 5 homicides per month in a given city. What
is the probability there will be at most 1 in a certain month?
Solution :
If
X
EX = 5. Since the expectation
P(X = 0) + P(X = 1) = e−5 + 5e−5 .
is the number of homicides, we are given that
for a Poisson is
Example 6.2.
λ,
then
λ = 5.
Therefore
Suppose on average there is one large earthquake per year in California.
What's the probability that next year there will be exactly 2 large earthquakes?
Solution : λ = EX = 1, so P(X = 2) = e−1 ( 21 ).
We have the following proposition connecting binomial and Poisson distributions.
Proposition 6.3 (Binomial approximation of Poisson distribution)
Xn is a binomial random variable with parameters n and pn
P(Xn = i) → P(Y = i), where Y is Poisson with parameter λ.
If
and
npn → λ,
then
84
6. SOME DISCRETE DISTRIBUTIONS
6.1 (Approximation of Poisson by binomials)
Note that by setting
pn := λ/n
for
n>λ
we can approximate the Poisson distribution with parameter
tions with parameters
n
and
λ
by binomial distribu-
pn .
This proposition shows that the Poisson distribution models binomials when the probability
of a success is small. The number of misprints on a page, the number of automobile accidents,
the number of people entering a store, etc. can all be modeled by a Poisson distribution.
Proof. For simplicity, let us suppose that
can use
λn = npn −−−→ λ.
n→∞
λ = npn
for
n > λ.
In the general case we
We write
P(Xn = i) =
n!
pi (1 − pn )n−i
i!(n − i)! n
i n−i
λ
n(n − 1) · · · (n − i + 1) λ
1−
=
i!
n
n
n
i
n(n − 1) · · · (n − i + 1) λ (1 − λ/n)
=
.
ni
i! (1 − λ/n)i
Observe that the following three limits exist
n(n − 1) · · · (n − i + 1)
−−−→ 1,
n→∞
ni
i
(1 − λ/n) −−−→ 1,
n→∞
(1 − λ/n) −−−→ e−λ ,
n
n→∞
which completes the proof.
In Section 2.2.3 we considered
k = 1, 2, . . . , n.
discrete uniform distributions
P(X = k) = n1
die (with n = 6),
with
This is the distribution of the number showing on a
for
for
example.
Geometric distribution
A random variable
X
has the geometric distribution with parameter
P(X = i) = (1 − p)
i−1
p
for
p, 0 < p < 1,
if
i = 1, 2, . . . .
Using a geometric series sum formula we see that
∞
X
∞
X
P(X = i) =
(1 − p)i−1 p =
i=1
In Bernoulli trials, if we let
i=1
X
1
p = 1.
1 − (1 − p)
X
X will be a geometric
X is the rst time we
be the rst time we have a success, then
random variable. For example, if we toss a coin over and over and
will have a geometric distribution. To see this, to have the rst success
6.1. EXAMPLES: BERNOULLI, BINOMIAL, POISSON, GEOMETRIC DISTRIBUTIONS
k th
occur on the
k−1
(1 − p)k−1 p.
trial, we have to have
success. The probability of that is
failures in the rst
k−1
85
trials and then a
Proposition 6.4
p, 0 < p < 1,
If X is a geometric random variable with parameter
then
1
EX = ,
p
1−p
,
p2
Var X =
FX (k) = P (X 6 k) = 1 − (1 − p)k .
Proof. We will use
∞
X
1
=
nrn−1
(1 − r)2
n=0
which we can show by dierentiating the formula for geometric series
1/(1 − r) =
P∞
Then
EX =
∞
X
i · P(X = i) =
i=1
∞
X
1
1
2 ·p = .
p
(1 − (1 − p))
i · (1 − p)i−1 p =
i=1
Then the variance
2 X
2
∞ 1
1
Var X = E (X − EX) = E X −
=
i−
· P(X = i)
p
p
i=1
2
To nd the variance we will use another sum. First
∞
X
r
nrn ,
=
(1 − r)2
n=0
which we can dierentiate to see that
∞
X
1+r
=
n2 rn−1 .
3
(1 − r)
n=1
Then
2
EX =
∞
X
2
i · P(X = i) =
i=1
∞
X
i2 · (1 − p)i−1 p =
i=1
2−p
(1 + (1 − p))
.
3 ·p =
p2
(1 − (1 − p))
Thus
2−p
Var X = EX − (EX) =
−
p2
2
2
2
1−p
1
=
.
p
p2
n=0
rn .
86
6. SOME DISCRETE DISTRIBUTIONS
The cumulative distribution function (CDF) can be found by using the geometric series sum
formula
∞
X
1 − FX (k) = P (X > k) =
P(X = i) =
i=k+1
∞
X
i−1
(1 − p)
i=k+1
(1 − p)k
p = (1 − p)k .
p=
1 − (1 − p)
Negative binomial distribution
A random variable
X
has
negative binomial distribution
P(X = n) =
with parameters
r
and
p
if
n−1 r
p (1 − p)n−r , n = r, r + 1, . . . .
r−1
A negative binomial represents the number of trials until r successes. To get the above
th
th
success in the n
trial, we must exactly have r − 1 successes in the
formula, to have the r
th
rst n − 1 trials and then a success in the n
trial.
Hypergeometric distribution
A random variable
X
has hypergeometric distribution with parameters
m, n
and
N
if
m
N −m
i
n−i
.
P(X = i) =
N
n
N balls, of which m are one
N −m are another, and we choose n balls at random without replacement,
the probability of having i balls of the rst color.
This comes up in sampling without replacement: if there are
color and the other
then
X
represents
Another model where the hypergeometric distribution comes up is the probability of a success
changes on each draw, since each draw decreases the population, in other words, when we
consider sampling without replacement from a nite population). Then
size,
m
is the number of success states in the population,
quantity drawn in each trial,
i
n
N
is the population
is the number of draws, that is,
is the number of observed successes.
6.2. FURTHER EXAMPLES AND APPLICATIONS
87
6.2. Further examples and applications
6.2.1. Bernoulli and binomial random variables.
Example 6.3.
A company prices its hurricane insurance using the following assumptions:
(i) In any calendar year, there can be at most one hurricane.
(ii) In any calendar year, the probability of a hurricane is 0.05.
(iii) The numbers of hurricanes in dierent calendar years are mutually independent.
Using the company's assumptions, nd the probability that there are fewer than 3 hurricanes
in a 20-year period.
Solution :
X the number of hurricanes
X ∼ Binom (20, 0.05), therefore
denote by
we see that
in a 20-year period. From the assumptions
P (X < 3) = P (X 6 2)
20
20
20
1
19
0
20
=
(0.05) (0.95) +
(0.05) (0.95) +
(0.05)2 (0.95)18
0
1
2
= 0.9245.
Example 6.4.
Phan has a
0.6 probability of making a free throw.
Suppose each free throw
is independent of the other. If he attempts 10 free throws, what is the probability that he
makes at least 2 of them?
Solution :
If
X ∼ Binom (10, 0.6),
then
P (X > 2) = 1 − P (X = 0) − P (X = 1)
10
10
0
10
=1−
(0.6) (0.4) −
(0.6)1 (0.4)9
0
1
= 0.998.
6.2.2. The Poisson distribution.
Recall that a Poisson distribution models well events
that have a low probability and the number of trials is high. For example, the probability of
a misprint is small and the number of words in a page is usually a relatively large number
compared to the number of misprints.
(1) The number of misprints on a random page of a book.
(2) The number of people in community that survive to age
100.
(3) The number of telephone numbers that are dialed in an average day.
(4) The number of customers entering post oce on an average day.
Example 6.5.
Levi receives an average of two texts every
3
minutes. If we assume that
the number of texts is Poisson distributed, what is the probability that he receives ve or
more texts in a
9-minute
period?
88
6. SOME DISCRETE DISTRIBUTIONS
Solution :
Let
X
be the number of texts in a
9−minute
period. Then
λ=3·2=6
and
P (X > 5) = 1 − P (X 6 4)
=1−
4
X
6n e−6
n=0
n!
= 1 − 0.285 = 0.715.
Example 6.6.
tation
λ.
Solution :
Let
X1 ,
...,
Xk
be independent Poisson random variables, each with expec-
What is the distribution of the random variable
The distribution of
Y
is Poisson with the expectation
use Proposition 6.3 and (6.1) to choose
pn = kλ1 /n = λ1 /m = λ/n
Y := X1 + ... + Xk ?
n = mk
λ = kλ.
To show this, we
Bernoulli random variables with parameter
to approximation the Poisson random variables.
them all together, the limit as
n → ∞
If we sum
gives us a Poisson distribution with expectation
lim npn = λ. However, we can re-arrange the same n = mk Bernoulli random variables
n→∞
in k groups, each group having m Bernoulli random variables. Then the limit gives us the
distribution of
X1 + ... + Xk .
This argument can be made rigorous, but this is beyond the
scope of this course. Note that we do not show that the we have convergence in distribution.
Example 6.7.
Let
λ1 , . . . , λk ,
X1 + ... + Xk ?
pectation
Solution :
X1 , . . . , X k
be independent Poisson random variables, each with ex-
respectively.
Y =
λ = λ1 + ... + λk . To show
this, we again use Proposition 6.3 and (6.1) with parameter pn = λ/n. If n is large, we can
separate these n Bernoulli random variables in k groups, each having ni ≈ λi n/λ Bernoulli
random variables. The result follows if lim ni /n = λi for each i = 1, ..., k .
The distribution of
Y
What is the distribution of the random variable
is Poisson with expectation
n→∞
This entire set-up, which is quite common, involves what is called
distributed Bernoulli random variables
Example 6.8.
independent identically
(i.i.d. Bernoulli r.v.).
Can we use binomial approximation to nd the mean and the variance of
a Poisson random variable?
Solution :
Yes, and this is really simple. Recall again from Proposition 6.3 and (6.1) that we
can approximate Poisson
where
pn = λ/n.
Y
with parameter
λ
by a binomial random variable
Each such a binomial random variable is a sum on
random variables with parameter
pn .
n→∞
n independent Bernoulli
Therefore
EY = lim npn = lim n
n→∞
λ
= λ,
n
λ
Var(Y ) = lim npn (1 − pn ) = lim n
n→∞
n→∞ n
Binom (n, pn ),
λ
1−
= λ.
n
6.2. FURTHER EXAMPLES AND APPLICATIONS
6.2.3. Table of distributions.
The following table summarizes the discrete distribu-
tions we have seen in this chapter.
N0 = N ∪ {0}
Here
stands for the set of positive integers, and
N
is the set of nonnegative integers.
PMF (k
Name
Notation
Parameters
Bernoulli
Bern(p)
p ∈ [0, 1]
1
k
Binomial
Binom(n, p)
n∈N
p ∈ [0, 1]
n
k
Poisson
Pois(λ)
λ>0
Geometric
Geo(p)
p ∈ (0, 1)
Negative
NBin(r, p)
r∈N
p ∈ (0, 1)
binomial
Hypergeometric
89
Hyp(N, m, n) N ∈ N0
n, m ∈ N0
∈ N0 )
pk (1 − p)1−k
pk (1 − p)n−k
E[X] Var(X)
p
p(1 − p)
np
np(1 − p)
k
e−λ λk!
λ
(
(1 − p)k−1 p, for k > 1,
1
p
0,
else.
(
k−1 r
p (1 − p)k−r , if k ≥ r, r
r−1
p
0,
else.
−m
(mk)(Nn−k
)
N
(n)
nm
N
λ
1−p
p2
r(1−p)
p2
nm(N −n)
m
(1−N
)
N (N −1)
90
6. SOME DISCRETE DISTRIBUTIONS
6.3. Exercises
Exercise 6.1.
A UConn student claims that she can distinguish Dairy Bar ice cream from
Friendly's ice cream. As a test, she is given ten samples of ice cream (each sample is either
from the Dairy Bar or Friendly's) and asked to identify each one. She is right eight times.
What is the probability that she would be right exactly eight times if she guessed randomly
for each sample?
Exercise 6.2.
A Pharmaceutical company conducted a study on a new drug that is sup-
posed to treat patients suering from a certain disease. The study concluded that the drug
did not help 25% of those who participated in the study. What is the probability that of 6
randomly selected patients, 4 will recover?
Exercise 6.3.
20% of all students are left-handed. A class of size 20 meets in a room with
18 right-handed desks and 5 left-handed desks. What is the probability that every student
will have a suitable desk?
Exercise 6.4.
A ball is drawn from an urn containing 4 blue and 5 red balls. After the
ball is drawn, it is replaced and another ball is drawn. Suppose this process is done
7
times.
(a) What is the probability that exactly 2 red balls were drawn in the 7 draws?
(b) What is the probability that at least 3 blue balls were drawn in the 7 draws?
Exercise 6.5.
The expected number of typos on a page of the new Harry Potter book is
0.2. What is the probability that the next page you read contains
(a) 0 typos?
(b) 2 or more typos?
(c) Explain what assumptions you used.
Exercise 6.6.
The monthly average number of car crashes in Storrs, CT is 3.5. What is
the probability that there will be
(a) at least 2 accidents in the next month?
(b) at most 1 accident in the next month?
(c) Explain what assumptions you used.
Exercise 6.7.
Suppose that, some time in a distant future, the average number of bur-
glaries in New York City in a week is 2.2.
Approximate the probability that there will
be
(a) no burglaries in the next week;
(b) at least 2 burglaries in the next week.
Exercise 6.8.
The number of accidents per working week in a particular shipyard is Poisson
distributed with mean
0.5.
Find the probability that:
(a) In a particular week there will be at least 2 accidents.
6.3. EXERCISES
91
(b) In a particular two week period there will be exactly 5 accidents.
(c) In a particular month (i.e. 4 week period) there will be exactly 2 accidents.
Exercise 6.9.
Jennifer is baking cookies. She mixes 400 raisins and 600 chocolate chips
(a) Find the probability that a randomly picked cookie will have three raisins in it.
(b) Find the probability that a randomly picked cookie will have at least one chocolate chip
in it.
(c) Find the probability that a randomly picked cookie will have no more than two bits in
it (a bit is either a raisin or a chocolate chip).
Exercise 6.10.
A roulette wheel has 38 numbers on it: the numbers 0 through 36 and a
00. Suppose that Lauren always bets that the outcome will be a number between 1 and 18
(including 1 and 18).
(a) What is the probability that Lauren will lose her rst 6 bets.
(b) What is the probability that Lauren will rst win on her sixth bet?
Exercise 6.11.
In the US, albinism occurs in about one in 17,000 births. Estimate the
probabilities no albino person, of at least one, or more than one albino at a football game with
5,000 attendants. Use the Poisson approximation to the binomial to estimate the probability.
Exercise 6.12.
An egg carton contains 20 eggs, of which 3 have a double yolk. To make a
pancake, 5 eggs from the carton are picked at random. What is the probability that at least
2 of them have a double yolk?
Exercise 6.13.
Around 30,000 couples married this year in CT. Approximate the proba-
bility that at least in one of these couples
(a) both partners have birthday on January 1st.
(b) both partners celebrate birthday in the same month.
Exercise 6.14.
A telecommunications company has discovered that users are three times
as likely to make two-minute calls as to make four-minute calls. The length of a typical call
(in minutes) has a Poisson distribution. Find the expected length (in minutes) of a typical
call.
92
6. SOME DISCRETE DISTRIBUTIONS
6.4. Selected solutions
Solution to Exercise 6.1:
This should be modeled using a binomial random variable
X,
since there is a sequence of trials with the same probability of success in each one. If
1
she guesses randomly for each sample, the probability that she will be right each time is .
2
Therefore,
8 2
1
45
10
1
= 10 .
P (X = 8) =
2
2
2
8
Solution to Exercise 6.2:
Solution to Exercise 6.3:
6
4
(0.75)4 (0.25)2
For each student to have the kind of desk he or she prefers, there
must be no more than 18 right-handed students and no more than 5 left-handed students, so
the number of left-handed students must be between 2 and 5 (inclusive). This means that
we want the probability that there will be 2, 3, 4, or 5 left-handed students. We use the
binomial distribution and get
5 i 20−i
X
20
1
4
i=2
i
5
5
.
Solution to Exercise 6.4(A):
2 5
7
5
4
2
9
9
Solution to Exercise 6.4(B):
P (X > 3) = 1 − P (X 6 2)
0 7 1 6 2 5
5
7
4
5
7
4
5
7
4
−
−
=1−
9
9
1
9
9
2
9
9
0
Solution to Exercise 6.5(A): e−0.2
Solution to Exercise 6.5(B): 1 − e−0.2 − 0.2e−0.2 = 1 − 1.2e−0.2 .
Solution to Exercise 6.5(C): Since each word has a small probability of being a typo, the
number of typos should be approximately Poisson distributed.
Solution to Exercise 6.6(A): 1 − e−3.5 − 3.5e−3.5 = 1 − 4.5e−3.5
Solution to Exercise 6.6(B): 4.5e−3.5
Solution to Exercise 6.6(C): Since each accident has a small probability it seems reasonable to suppose that the number of car accidents is approximately Poisson distributed.
Solution to Exercise 6.7(A): e−2.2
Solution to Exercise 6.7(B): 1 − e−2.2 − 2.2e−2.2 = 1 − 3.2e−2.2 .
6.4. SELECTED SOLUTIONS
93
Solution to Exercise 6.8(A): We have
P (X > 2) = 1 − P (X 6 1) = 1 − e−0.5
(0.5)0
(0.5)1
− e−0.5
.
0!
1!
Solution to Exercise 6.8(B): In two weeks the average number of accidents will be λ =
0.5 + 0.5 = 1.
Then
5
P (X = 5) = e−1 15! .
Solution to Exercise 6.8(C): In a 4 week period the average number of accidents will be
λ = 4 · (0.5) = 2.
Then
2
P (X = 2) = e−2 22! .
Solution to Exercise 6.9(A):
number of raisins per cookie is
3
−0.8 (0.8)
which is e
≈ 0.0383.
3!
This calls for a Poisson random variable
0.8,
Solution to Exercise 6.9(B):
so we take this as our
which is
1.2, so we take this
0
1 − P (C = 0) = 1 − e−1.2 (1.2)
≈ 0.6988.
0!
Solution to Exercise 6.9(C):
The average
This calls for a Poisson random variable
number of chocolate chips per cookie is
P (C > 1),
λ
R.
as our
λ.
C.
P(R = 3),
The average
This calls for a Poisson random variable
B.
The average
0.8 + 1.2 = 2, so we take this as our λ. We are asking for
0
1
2
P (B = 0) + P (B = 1) + P (B = 2) = e−2 20! + e−2 21! + e−2 22! ≈ .6767.
number of bits per cookie is
P (B 6 2),
which is
Solution to Exercise 6.10(A): 1 − 18
38
6
Solution to Exercise 6.10(B): 1 − 18
38
5
18
38
Solution to Exercise 6.11 Let X denote the number of albinos at the game.
We have that
X ∼ Binom(5000, p) with p = 1/17000 ≈ 0.00029. The binomial distribution gives us
5000
5000
P(X = 0) = 16999
≈ 0.745
P(X > 1) = 1 − P(X = 0) = 1 − 16999
≈ 0.255
17000
17000
P(X > 1) = P(X > 1) − P(X = 1) =


=1−
16999 5000
17000
5000

−


1
16999 4999
17000
Approximating the distribution of
X
1
1
17000
≈ 0.035633
by a Poisson with parameter
λ=
5000
17000
=
5
gives
17
5
P(Y = 0) = exp − 17
≈ 0.745
5
P(Y > 1) = 1 − P(Y = 0) = 1 − exp − 17
≈ 0.255
5
5
5
P(Y > 1) = P(Y > 1) − P(Y = 1) = 1 − exp − 17
− exp − 17
≈ 0.035638
17
Solution to Exercise 6.12:
Let
X
be the random variable that denotes the number of
X ∼ Hyp(20, 3, 5) and we
17
17
3
3
·
·
P(X > 2) = P(X = 2) + P(X = 3) = 2 203 + 3 202 .
eggs with double yolk in the set of chosen 5. Then
5
Solution to Exercise 6.13:
We will use Poisson approximation.
5
have that
94
6. SOME DISCRETE DISTRIBUTIONS
1
. If X
3652
denotes the number of married couples where this is the case, we can approximate the
−2
distribution of X by a Poisson with parameter λ = 30, 000 · 365
≈ 0.2251. Hence,
−0.2251
P(X > 1) = 1 − P(X = 0) = 1 − e
.
(a) The probability that both partners have birthday on January 1st is
p =
(b) In this case, the probability of both partners celebrating birthday in the same month
1/12 and therefore we approximate the distribution by a Poisson
λ = 30, 000/12 = 2500. Thus, P(X > 1) = 1 − P(X = 0) = 1 − e−2500 .
is
Solution to Exercise 6.14:
tion, X ∼ Pois(λ) for
Let
X
denote the duration (in minutes) of a call. By assump-
λ > 0, so that the expected duration
P(X = 2) = 3P(X = 4), which means
some parameter
we know that
λ2
λ4
= 3e−λ .
2!
4!
hence E[X] = λ = 2.
e−λ
From here we deduce that
λ2 = 4
with parameter
and
of a call is
Part 2
Continuous random variables
CHAPTER 7
Continuous distributions
7.1. Basic theory
7.1.1. Denition, PDF, CDF.
variable.
Denition (Continuous random variables)
X is said to have
f = fX such that
A random variable
negative function
a
continuous distribution
if there exists a non-
b
P(a 6 X 6 b) =
f (x)dx
a
for every
a
and
b.
f
The function
is called the
density function
for
X
or the
PDF
for
X.
More precisely, such an
∞
f (x)dx = P(−∞ <
−∞
Example
7.1.
∞
−∞
f (x)dx = 1
X is said to have an absolutely continuous
a distribution. Note that
X < ∞) = 1. In particular, P(X = a) = a f (x)dx = 0 for every a.
Suppose we are given that
and
f (x) = c/x3
∞
f (x)dx = c
c
−∞
we have
∞
1
for
x>1
and
0
otherwise. Since
1
c
dx = ,
3
x
2
c = 2.
PMF or PDF?
Probability mass function (PMF) and (probability) density function (PDF) are two
names for the same notion in the case of discrete random variables. We say
PDF
simply a
only for
density function
for a general random variable, and we use
discrete random variables.
Denition (Cumulative distribution function (CDF))
The
distribution function
of
X
is dened as
y
F (y) = FX (y) := P(−∞ < X 6 y) =
It is also called the
cumulative distribution function
97
f (x)dx.
−∞
(CDF) of
X.
PMF
or
98
7. CONTINUOUS DISTRIBUTIONS
We can dene CDF for any random variable, not just continuous ones, by setting
P(X 6 y).
F (y) :=
Recall that we introduced it in Denition 5.3 for discrete random variables. In
that case it is not particularly useful, although it does serve to unify discrete and continuous
random variables.
provided
f
In the continuous case, the fundamental theorem of calculus tells us,
satises some conditions, that
f (y) = F 0 (y) .
By analogy with the discrete case, we dene the expectation of a continuous random variable.
7.1.2. Expectation, discrete approximation to continuous random variables.
Denition (Expectation)
For a continuous random variable
tation
X
with the density function
by
f
we dene its
expec-
∞
xf (x)dx
EX =
−∞
if this integral is absolutely convergent. In this case we call
X integrable.
Recall that this integral is absolutely convergent if
∞
|x|f (x)dx < ∞.
−∞
In the example above,
∞
EX =
1
2
x 3 dx = 2
x
∞
x−2 dx = 2.
1
Later in Example 10.1 we will see that a continuous random variable with Cauchy distribution
has innite expectation.
Proposition 7.1 (Discrete approximation to continuous random variables)
Suppose
X
is a nonnegative continuous random variable with a nite expectation.
{Xn }∞
n=1 such that
Then there is a sequence of discrete random variables
EXn −−−→ EX.
n→∞
Proof. First observe that if a continuous random variable
density
f (x) = 0 x < 0.
In particular,
F (y) = 0
for
y 6 0,
X
is nonnegative, then its
thought the latter is not needed
for our proof. Thus for such a random variable
∞
EX =
xf (x)dx.
0
n ∈ N, then we dene Xn (ω) to be k/2n if k/2n 6 X(ω) < (k+1)/2n , for k ∈ N∪{0}.
−n
This means that we are approximating X from below by the largest multiple of 2
that is
still below the value of X . Each Xn is discrete, and Xn increase to X for each ω ∈ S .
Suppose
7.1. BASIC THEORY
Consider the sequence
{EXn }∞
n=1 .
99
This sequence is an increasing sequence of positive num-
bers, and therefore it has a limit, possibly innite. We want to show that it is nite and it
is equal to
EX .
We have
∞
X
k
k
EXn =
P Xn = n
2n
2
k=1
∞
X
k
k+1
k
=
P n 6X<
2n
2
2n
k=1
(k+1)/2n
∞
X
k
=
f (x)dx
n
2
n
k/2
k=1
=
∞ X
k=1
If
x ∈ [k/2n , (k + 1)/2n ),
then
x
k/2n
diers from
k
f (x)dx.
2n
k/2n
(k+1)/2n
by at most
(k+1)/2n
xf (x)dx −
06
(k+1)/2n
k/2n
(k+1)/2n
=
k/2n
k/2n
1/2n ,
and therefore
k
f (x)dx
2n
(k+1)/2n
k
1
f (x)dx
x − n f (x)dx 6 n
2
2 k/2n
Note that
∞ X
k=1
(k+1)/2n
∞
xf (x)dx =
k/2n
xf (x)dx
0
and
(k+1)/2n
∞
n
∞
∞ X
1
1
1 X (k+1)/2
1
f (x)dx = n
f (x)dx = n
f (x)dx = n .
n
2 k/2n
2 k=1 k/2n
2 0
2
k=1
Therefore
100
7. CONTINUOUS DISTRIBUTIONS
∞
xf (x)dx −
0 6 EX − EXn =
0
=
=
6
∞ X
k=1
xf (x)dx −
k/2n
k=1
∞
X
k=1
∞
X
(k+1)/2n
∞ X
k=1
(k+1)/2n
∞ X
k=1
(k+1)/2n
k/2n
(k+1)/2n
k/2n
(k+1)/2n
k/2n
1
2n
k/2n
(k+1)/2n
f (x)dx =
k/2n
k
f (x)dx
2n
!
k
f (x)dx
2n
xf (x)dx −
k
f (x)dx
2n
1
−−→ 0.
2n n→0
We will not prove the following, but it is an interesting exercise: if
discrete random variables that increase up to
X , then limm→∞ EXm
Xm
is
any
sequence of
will have the same value
EX .
This fact is useful to show linearity, if
expectations, then we can take
up to
Y.
Then
Xm + Ym
Xm
X
and
Y
are positive random variables with nite
X and Ym discrete
X + Y , so we have
discrete increasing up to
is discrete and increases up to
increasing
E(X + Y ) = lim E(Xm + Ym )
m→∞
= lim EXm + lim EYm = EX + EY.
m→∞
m→∞
Note that we can not easily use the approximations to
previous proof to use in this argument, since
Xm + Ym
X, Y
and
X+Y
we used in the
might not be an approximation of the
same kind.
If
X
is not necessarily positive, we can show a similar result; we will not do the details.
Similarly to the discrete case, we have
Proposition 7.2
Suppose
X
is a continuous random variable with density
fX
and
g
is a real-valued
function, then
∞
Eg(X) =
g(x)f (x)dx
−∞
as long as the expectation of the random variable
g (X)
makes sense.
As in the discrete case, this allows us to dene moments, and in particular the
Var X := E[X − EX]2 .
As an example of these calculations, let us look at the uniform distribution.
variance
7.1. BASIC THEORY
101
Uniform distribution
We say that a random variable
if
a6x6b
X
has a
uniform distribution
on
[a, b]
if
and 0 otherwise.
To calculate the expectation of
X
∞
b
x
xfX (x)dx =
EX =
−∞
a
1
dx
b−a
b
1
x dx
b−a a
1 b 2 a2 a + b
−
=
.
=
b−a 2
2
2
=
This is what one would expect. To calculate the variance, we rst calculate
∞
b
x2
2
2
x fX (x)dx =
EX =
−∞
a
1
a2 + ab + b2
dx =
.
b−a
3
We then do some algebra to obtain
Var X = EX 2 − (EX)2 =
(b − a)2
.
12
fX (x) =
1
b−a
102
7. CONTINUOUS DISTRIBUTIONS
7.2. Further examples and applications
Example 7.2.
Suppose
X
has the following p.d.f.
(
f (x) =
Find the CDF of
Solution :
X,
we have
that is, nd
FX (x) = 0
if
FX (x).
x61
2
x3
x>1
x < 1.
0
Use the CDF to nd
and will need to compute
x
FX (x) = P (X 6 x) =
1
x > 1.
when
P (3 6 X 6 4).
2
1
dy = 1 − 2
3
y
x
We can use this formula to nd the following probability
P (3 6 X 6 4) = P (X 6 4) − P (X < 3)
1
1
7
= FX (4) − FX (3) = 1 − 2 − 1 − 2 =
.
4
3
144
Example 7.3.
Suppose
X
has density
(
2x 0 6 x 6 1
fX (x) =
.
0
otherwise
Find
EX .
Solution :
we have that
E [X] =
xfX (x)dx =
0
Example 7.4.
The density of
X
2
x · 2x dx = .
3
is given by
(
fX (x) =
Find
1
1
2
if
0
otherwise
06x62
.
E eX .
Solution :
using Proposition 7.2 with
g(x) = ex
2
EeX =
0
Example 7.5.
Suppose
X
we have
1
1 2
ex · dx =
e −1 .
2
2
has density
(
2x 0 6 x 6 1
f (x) =
.
0
otherwise
7.2. FURTHER EXAMPLES AND APPLICATIONS
Find
103
Var(X).
Solution :
in Example 7.3 we found
E X
2
E [X] =
2
. Now
3
1
x · 2xdx = 2
=
0
0
Thus
1
Var(X) = −
2
Example 7.6.
Suppose
X
1
2
1
x3 dx = .
2
2
2
1
= .
3
18
has density
(
ax + b 0 6 x 6 1
f (x) =
.
0
otherwise
and that
Solution :
E [X 2 ] =
1
. Find the values of
6
We need to use the fact that
us
a
and
∞
−∞
b.
f (x)dx = 1
1
1=
(ax + b) dx =
0
and the second one give us
1
=
6
x2 (ax + b) dx =
a b
+ .
4 3
Solving these equations gives us
a = −2,
and
E [X 2 ] =
a
+ b,
2
1
0
and
b = 2.
1
. The rst one gives
6
104
7. CONTINUOUS DISTRIBUTIONS
7.3. Exercises
Exercise 7.1.
Let
X
be a random variable with probability density function
(
cx (5 − x) 0 6 x 6 5,
f (x) =
0
otherwise.
(A) What is the value of
c?
X?
P (2 6 X ≤ 3).
(B) What is the cumulative distribution function of
(D) What is
(E) What is
That is, nd
FX (x) = P (X 6 x).
E [X]?
Var(X)?
Exercise 7.2.
UConn students have designed the new U-phone.
X
that the lifetime of a U-Phone is given by the random variable
They have determined
(measured in hours), with
probability density function
(
f (x) =
10
x2
x > 10,
x ≤ 10.
0
(A) Find the probability that the u-phone will last more than
(B) What is the cumulative distribution function of
Exercise 7.3.
Suppose the random variable
f (x) =
That is, nd
X
2
x2
0
has a density function
x > 2,
x 6 2.
An insurance company insures a large number of homes. The insured value,
of a randomly selected home is assumed to follow a distribution with density function
(
f (x) =
3
x4
x > 1,
0
otherwise.
Given that a randomly selected home is insured for at least
that it is insured for less than
Exercise 7.5.
If
FX (x) = P (X 6 x).
E [X].
Exercise 7.4.
X,
hours.
P (X > 35)?
(
Compute
X?
20
E [X] =
2.
X is given by
(
a + bx2 0 6 x 6 1,
f (x) =
0
otherwise.
The density function of
7
, nd the values of
10
a
and
b.
1.5,
calculate the probability
7.3. EXERCISES
Exercise 7.6.
Let
X
be a random variable with density function
(
f (x) =
Suppose that
E [X] = 6 Var(X).
Exercise 7.7.
105
1
a−1
0
1 < x < a,
otherwise.
Find the value of
a.
Suppose you order a pizza from your favorite pizzeria at 7:00 pm, knowing
that the time it takes for your pizza to be ready is uniformly distributed between 7:00 pm
and 7:30 pm.
(A) What is the probability that you will have to wait longer than
10
minutes for your
pizza?
(B) If at 7:15pm, the pizza has not yet arrived, what is the probability that you will have
to wait at least an additional
Exercise 7.8.
10
minutes?
X
of a machine part has a continuous distribution
on the interval (0, 10) with probability density function fX (x), where fX (x) is proportional
x
to
on the interval. The reparation costs of this part are modeled by a random variable Y
5
2
that is given by Y = 3X . Compute the expected cost of reparation of the machine part.
Exercise 7.9.
and
10 : 20,
A bus arrives at some (random) time uniformly distributed between
and you arrive at a bus stop at
10 : 00
10 : 05.
(A) What is the probability that you have to wait at least
(B) What is the probability that you have to wait at least
5 minutes until the bus comes?
5 minutes, given that when you
arrive today to the station the bus was not there yet (you are lucky today)?
Exercise∗ 7.1.
For a continuous random variable
X
with nite rst and second moments
prove that
E (aX + b) = aEX + b,
Var (aX + b) = a2 Var X.
for any
a, b ∈ R.
Exercise∗ 7.2.
Let
X
be a continuous random variable with probability density function
x
1
fX (x) = xe− 2 1[0,∞) (x) ,
4
where the indicator function is dened as
1[0,∞) (x) =




1, 0 6 x < ∞;


 0,
Check that
fX
otherwise.
is a valid probability density function, and nd
E (X)
if it exists.
106
7. CONTINUOUS DISTRIBUTIONS
Exercise∗ 7.3.
Let
X
be a continuous random variable with probability density function
fX (x) =
4 ln x
1[1,∞) (x) ,
x3
where the indicator function is dened as
1[1,∞) (x) =




1, 1 6 x < ∞;


 0,
Check that
fX
otherwise.
is a valid probability density function, and nd
E (X)
if it exists.
7.4. SELECTED SOLUTIONS
107
7.4. Selected solutions
Solution to Exercise 7.1(A): We must have that
∞
f (x)dx = 1,
−∞
5
5
5x2 x3
1=
cx(5 − x)dx = c
−
2
3
0
0
thus
c = 6/125.
and so we must have that
Solution to Exercise 7.1(B): We have that
x
f (y)dy
FX (x) = P (X 6 x) =
−∞
x
6
6
=
y (5 − y) dx =
125
0 125
2
3
6
5x
x
=
−
.
125
2
3
5y 2 y 3
−
2
3
x
0
Solution to Exercise 7.1(C): We have
P (2 6 X 6 3) = P (X 6 3) − P (X < 2)
6
5 · 32 33
5 · 22 23
6
=
−
−
−
125
2
3
125
2
3
= 0.296.
Solution to Exercise 7.1(D): we have
∞
E [X] =
5
x·
xfX (x)dx =
−∞
0
6
x(5 − x)dx
125
= 2.5.
Solution to Exercise 7.1(E): We need to rst compute
E X
2
∞
5
2
=
x2 ·
x fX (x)dx =
−∞
0
6
x(5 − x)dx
125
= 7.5.
Then
Var(X) = E X 2 − (E [X])2 = 7.5 − (2.5)2 = 1.25.
Solution to Exercise 7.2(A): We have
∞
20
10
1
dx = .
2
x
2
Solution to Exercise 7.2(B): We have
x
F (x) = P(X 6 x) =
10
for
x > 10,
and
F (x) = 0
for
x < 10.
10
10
dy = 1 −
2
y
x
108
7. CONTINUOUS DISTRIBUTIONS
Solution to Exercise 7.2(C): We have
P (X > 35) = 1 − P (X < 35) = 1 − FX (35)
10
10
=1− 1−
= .
35
35
Solution to Exercise 7.3: +∞
Solution to Exercise 7.4:
Solution to Exercise 7.5:
37
.
64
we need to use the fact that
The rst one gives us
1
0
7
=
10
1
0
−∞
f (x)dx = 1
and
E [X] =
7
.
10
b
a + bx2 dx = a +
3
1=
and the second one gives
∞
a b
x a + bx2 dx = + .
2 4
Solving these equations gives
1
a= ,
5
Solution to Exercise 7.6:
and
b=
12
.
5
Note that
EX =
1
a
x
1
1
dx = a + .
a−1
2
2
Also
Var(X) = EX 2 − (EX)2
then we need
a
2
EX =
1
1
1
1
x2
dx = a2 + a + .
a−1
3
3
3
Then
2
1 2 1
1
1
1
Var(X) =
a + a+
a+
−
3
3
3
2
2
1
1
1
= a2 − a + .
12
6
12
E [X] = 6 Var(X), we simplify and get 12 a2 − 32 a = 0, which
Then, using
gives us
a = 3.
Another way to solve this problem is to note that, for the uniform distribution on [a, b],
(a−b)2
(a−1)2
a+b
the mean is
and the variance is
. This gives us an equation 6
= a+1
. Hence
2
12
12
2
2
(a − 1) = a + 1, which implies a = 3.
Solution to Exercise 7.7(A): Note that X
is uniformly distributed over
(0, 30).
Then
(0, 30).
Then
2
P(X > 10) = .
3
Solution to Exercise 7.7(B): Note that X
P(X > 25 | X > 15) =
is uniformly distributed over
P (X > 25)
5/30
=
= 1/3.
P(X > 15)
15/30
7.4. SELECTED SOLUTIONS
Solution to Exercise 7.8:
First of all we need to nd the PDF of
(
f (x) =
Since
we have
c=
109
cx
5
0
10
X.
So far we know that
0 6 x 6 10,
otherwise.
x
c dx = 10c,
5
0
1
. Now, applying Proposition 7.2 we get
10
10
3 3
EY =
0
50
x dx = 150.
Solution to Exercise 7.9(A):
The probability that you have to wait at least 5 minutes
1
1
until the bus comes is . Note that with probability
you have to wait less than 5 minutes,
2
4
1